The first excited vibrational energy level of diatomic chlo- rine (Cl2) is 558 cm^-1 above the ground state. Wave- numbers, the units in which vibrational frequencies are usually recorded, are effectively units of energy, with 1 cm ^-1 = 1.986445 X 10^-23 J. If every vibrational energy level is equally spaced, and has a degeneracy of 1, sum over the lowest 4 vibrational levels to obtain a vibrational partition function for chlorine.

Required:
Determine the populations of each of the four levels at 298 K and the average molar vibrational energy (Em.vib) for chlorine at 298 K.

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codiepienagoya codiepienagoya · Answer 1 · 2021-05-25T12:09:03+02:00

Answer:

The answer is "0.0000190 and 2.7 J".

Explanation:

[tex]\to P_4=\frac{e^{-\beta}(4+\frac{1}{2}) hev}{2vb}\\\\[/tex]

[tex]=\frac{e^{-9(1.35)}}{0.278v}\\\\=\frac{e^{-12.5}}{0.278}\\\\=\frac{0.000005285}{0.278}\\\\=0.0000190[/tex]

Given:

[tex]h=6.626 \times 10^{-34}\\\\c=3\times 10^{10}\\\\v=558\ cm^{-1}\\\\K=1.38 \times 10^{-23}\\\\ T=298\\\\[/tex]

[tex]E=\frac{hcv}{KT}\\\\[/tex]

by putting the value into the above formula so, the value is 2.7 J